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EE451/551:Digital Control

Chapter 5: Control System Design Using Analog Prototypes

Root Locus Method• The root locus method provides a quick means ofThe root locus method provides a quick means of

predicting the closed‐loop behaviour of a system based on its open‐loop poles and zeros

• The method is based on the properties of the closed‐loop characteristic equation

1 ( ) 0where is a design gain and the loop gain ( ) is

KL sK L s

+ =

( )

g g p g ( )zn

is z−∏

( )1 ( )p

in

j

L ss p

==−∏( )

1

where and are the open-loop zeros and poles

jj

i jz p=∏

Root Locus Method• It is desired to find the location of the closed‐loop polesIt is desired to find the location of the closed loop poles

(root loci) as the gain K varies between zero and infinity

• Since the pole locations are related to the system time p yresponse, the Root Locus Method can be used as a design tool to determine the desired gain K

• The complex expression

1 ( ) 0KL s+ =( )can be written as two real equalities know as:1 ( ) 1K L s =The magnitude condition

( )1. ( ) 1

2. ( ) 2 1 180 ,

K L s

L s m

=

∠ = ± +

The magnitude condition

The angle condition for 0,1, 2,m =

Root Loci Rules1 Th b f t l b h i l t1. The number of root locus branches is equal to ,

the number of open-loop poles of ( )pn

L s2. The root locus branches start at the open-loop poles, and end the the open - loop zeros (or infinity)p p ( y)3. The real axis root loci have an odd number of poles

plus zeros to their right on the real axis plus zeros to their right on the real axis4. The branches going to infinity asymptotically approach the straight lines defined by the angle

( )2 1 1800 1 2

mθ

+± ( )

, 0,1, 2, ap z

mn n

θ = ± =−

with real axis interceptp zn n

1 1

p z

j ij i

a

p zσ = =

−=∑ ∑

5. Breakaway points (points of departure from the real axis)p zn n−

correspond to local maxima of K( ), based on the CE,

σ whereas break-in points (points of arrival at the real axis)

correspond to local minima of K( )6. The angle of departure from a complex pole is given bynp

σ

( )1

g p p p g y

180p

nn

n j

p

p p p−

− ∠ − + ∠∑ ( )zn

n iz−∑( )1

jj=∑

1

The angle of arrival to a complex zero is defined similarlyi=∑

Example Root Locus Plot

( )pG s

To create the RL plot for the above system apply the rules:To create the RL plot for the above system, apply the rules: 1. The number of root locus branches is equal to

2 R l b h h

=2

l lpn

2. Root locus branches start at the oi.e.,

pen0, 1, and end at infin

-loop poles, s = − ity (since no finite zeros)3. Real axis root loci have an odd number of poles plus zeros to their right on the real axisg root loci on the real axis between the two real p s ole⇒

4. Branches going to infinity asymptotically approachthe straight lines defined by the angle the straight lines defined by the angle

180 90aθ = ± = ±

with real axis intercept2a

1 1 1 1

p zn n

j ij ip z

= =

−−∑ ∑

1 1 2

0

j ia

p zn nσ = = = −

− − 2

5. Breakaway points (points of departure from the real axis)d l l i f K( ) b d h CE correspond to local maxima of K( ), based on the CE,

whereas break-in points (points of arrival at the real axis)σ

correspond to lo

( ) ( )2( ) 1, or 1

cal minima of K( )

KL K

σ

σ σ σ σ σ⇒ = − = − + = − +( ) ( ) ( ) 1, or 1

1 2 1 02b

KL K

dKd

σ σ σ σ σ

σ σ

⇒ + +

⇒ − = + = ⇒ = −

( )2

21 The gain at breakaway is

bd

K

σ

σ σ⇒ = − + =( )1 2

g

6. Does not apply, since all poles real!

y4bσ =−

>> num_s=[1];den_s=conv([1 0],[1 1]);>> G s=tf(num s den s)>> G_s=tf(num_s,den_s)

Transfer function:1

‐‐‐‐‐‐‐s^2 + s

>> rlocus(G s) rlocus(G_s)

Questions?

In‐class Exercises• Sketch the RL plots for the following loop gains:

( )( )

Sketch the RL plots for the following loop gains:11. ( )

1 3L s

s s=

+ +( )( )

( )( )( )

1 312. ( )

1 3 5

s s

L ss s s

+ +

=+ + +( )( )( )( )

( )( )

1 3 5

53. ( )

1 3

s s s

sL s

+ + +

+=

+ +( )( )1 3s s+ +

Matlab Verification

[ ] d ([ ] [ ])>> num_s=[1];den_s=conv([1 1],[1 3]);>> G_s=tf(num_s,den_s)Transfer function:

1‐‐‐‐‐‐‐‐‐‐‐‐‐s^2 + 4 s + 3>> rlocus(G_s)

>> num_s=[1];>>den_s=conv(conv([1 1],[1 3]),[1 5]);>> G_s=tf(num_s,den_s)Transfer function:

11‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐s^3 + 9 s^2 + 23 s + 15>> rlocus(G_s); axis([‐8 4 ‐6 6])

>> num_s=[1 5];>>den s=conv([1 1] [1 3]);>>den_s=conv([1 1],[1 3]);>> G_s=tf(num_s,den_s) Transfer function:

s + 5‐‐‐‐‐‐‐‐‐‐‐‐‐s^2 + 4 s + 3 >> rlocus(G_s)

Soln. for HW P2.9(d)• In Chapter 2 we showed that once the residues ofA• In Chapter 2, we showed that once the residues of

( ) are known, then ( ) is converted

in

i

AA zF z F z =∑

1, ( )

to ( ) using the transform table and propertiesi iz z p

f k= −∑

• If ( ) contains complex conjugate poles at F z p θ∠± p ,

h i PFE ill h f h fthen its PFE will have terms of the form:

( ) +Az A zF z∗

= +

( )

( ) +

( ) + 2 coskk kA

F zz p z p

f k Ap A p A p kθ θ

∗

∗ ∗

= +− −

= + = + +( )( ) + 2 cos p Af k Ap A p A p kθ θ+ + +

% See script P2_9d.m on class web site

Soln. for HW P2.9(d)>> num_z=[1 ‐0.1];den_z=[1 0.04 .25]; >> F_z=tf(num_z,den_z,‐1)Transfer function:

z 0 1z ‐ 0.1‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐z^2 + 0.04 z + 0.25Sampling time: unspecified

>> [A,p,K]=residue(num_z,conv(den_z,[1 0])) % Compute residues for F_z over zA = 0.2000 ‐ 1.0088i

0.2000 + 1.0088i‐0 40000.4000

p = ‐0.0200 + 0.4996i‐0.0200 ‐ 0.4996i0

[]K = []>> k=0:5; >> f=A(3)*(k<1)+2*abs(A(1))*abs(p(1)).^k.*cos(angle(p(1))*k+angle(A(1))); % notes soln.>> fs=‐0.4*(k<1)+2.057*(0.5).^k.*sin(1.611*k+0.196); % text soln.( ) ( ) ( );>> impulse(F_z,k); axis([0 5 ‐0.25 1]) % numerical soln. >> hold;stem(k,f,'go');stem(k,fs,'rx')

Soln. for HW P2.9(d)

Time‐Domain Design Specifications• The following are common time domain specifications givenThe following are common time domain specifications given

for closed-loop system responses, typ. to a step input:1 Time constant : Time to reach 63% of the final valueτ1. Time constant, : Time to reach 63% of the final value2. Rise time, : Time to go frrT

τom 10 to 90% of the final value

3 Time delay : Time to reach 50% of the final valueT3. Time delay, : Time to reach 50% of the final value Peak value Final value4. Percent overshoot, 100%

Fi l l

dT

PO −= ×

Final value5. Peak time, : Time to first peak of an oscpT illatory response

6 S ttli ti Ti ft hi h ill tT6. Settling time, : Time after which an oscillatory responseremains within a percentage (typ. 2 or 5%) of the final value7 S d ( ) S Ch 3

sT

7. Steady-state error, ( ) : See Chapter 3 notes sse e= ∞

Frequency‐Domain Design Specifications• Criteria 1 2 3 6 and 7 can be applied to first-order closed-loop•

( )

Criteria 1, 2, 3, 6 and 7 can be applied to first-order closed-loopsystems of the form:

1KK Kτ( )

( )( )

1 ( )

1hil ll i i b li d d d l d l

t atcl

KKa KH s e Kaes a s

τττ τ

− −= = =+ +

while all criteria can be applied to second-order closed-loopsystems of the form:

( ) ( )2 2

2 2 ( ) sin

2atn n

cl ddn n

K KH s e ts s

ω ω ωωζω ω

−=+ +

2 2

where 0< <1 is assumed, i.e., complex conjugate pole pairs

located at 1 and 1s a j

ζ

ζω ω ζ ω ω ω ζ= − ± − = − ± = −located at 1 , and 1n n d d ns a jζω ω ζ ω ω ω ζ= ± = ± =

Relating Time and Frequency Domain Specifications1 11 Time constant τ1. Time constant,

0.8 2.52 Rise time note: this is a linear appro

na

T

τζωζ

= =

+≅2. Rise time, , note: this is a linear approx.

1 0.73 Ti d l t thi i li

rn

T

T

ζω

ζ

≅

+≅

21

3. Time delay, , note: this is a linear approx. dn

Tπζζ

ζω

−

≅

214. Percent overshoot, 100%

5 Peak time

PO e

T

ζ−= ×π

=5. Peak time, T

4 36 Settling time or

pd

T T

ω

= =2% 5%6. Settling time, or

7. Steady-state error, ( ) See Chapter 3

s sn n

ss

T T

e eζω ζω

= =

= ∞ =

Design Example Relating Specifications• Design a closed loop system response that approx.g p y p pp

a second order system with peak time, 0.89s,pd

T πω

= =

2%4and settling time 1.14 s. From the specs.,

h h i

sn

Tζω

= =

we have the constraints:

1π

2

40.89 and 1.14ζωζ

= =

•

1nω2

Solving these eqns. simultaneously yields 5 and 0.707

n

n

ζωζ

ω ζ

−

≈ ≈

2

desired closed-loop poles are located at:

1 3 54 3 54

n

s j j jζω ω ζ α ω

⇒

= − ± − = ± = − ±2 2( ) 2

1 3.54 3.54

the CP is given by cl n n

cl n n d

s s s

s j j j

ζω ω

ζω ω ζ α ω

∆ = + +

= ± = ± = ±

⇒ 2 7.08 25.06s s= + +

Matlab Solution>> clear;format short>> clear;format short>> [zeta,wn]=solve('pi/wn/sqrt(1‐zeta^2)=0.89','4/zeta/wn=1.14');>> wn=eval(wn),zeta=eval(zeta)

wn =

4.9771

zeta =zeta

0.7050>> s eta* n+j* n*sqrt(1 eta^2)>> s=‐zeta*wn+j*wn*sqrt(1‐zeta^2)

s =‐3.5088 + 3.5299i

Design Example Relating Specifications• If we assume a unit dc gain, i.e., 1, this implies aK =

2

closed-loop TF of the form:25( ) nKH s ω

( ) ( )2 2 2( )

2 7.07 25

Using Matlab to plot the closed-loop step response yields:

ncl

n n

H ss s s sζω ω

= =+ + + +

Using Matlab to plot the closed loop step response yields:>> Hcl_s=tf([25],[1 7.07 25])

Transfer function:25

‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐s^2 + 7.07 s + 25s 2 + 7.07 s + 25

>> step(Hcl_s)

Common Control Topologies• Some common controller topologies are illustrated below:

Cascade compensator (controller):

11

1 1

( ) ( )( )

( ) ( ) ( ) ( )c p

clc p c p

N s N sH s

N s N s D s D s=

+

p ( )

1

1

( )( )

cl

cl

N ss

=∆

11

1

( )( ) ( )c

cc

N sG s D s= ( )( ) ( )p

pp

N sG s D s=Focus of the text Perform RL analysis to createdesired Characteristic Poly.

Feedback compensator (controller):

22

2 2

( ) ( )( )

( ) ( ) ( ) ( )c p

clc p c p

D s N sH s

N s N s D s D s=

+

p ( )

22

( )( ) ( )

cc

N sG s D s=

2

2

( )( )

cl

cl

N ss

=∆

1 2

1 2

Note: If ( ) ( ),then ( ) ( )

c c

cl cl

G s G ss s

≡∆ = ∆

2( )cD s

12 ( )cG s−

2( )cG s

Note: The following cascade‐based controller is equivalent to the feedback controller above:

2c 2( )c

Common Control TopologiesMixed compensator (controller) consisting of an inner and outer loop:

11

( )( ) ( )

cc

N sG s D=1 2

3( ) ( )( )

( ) ( ) ( ) ( )c cl

clN s N sH s

N s N s D s s=

+ ∆11

( ) ( )ccD s

22

2

( )( ) ( )

cc

c

N sG s D s=

1 2 1 2

3

3

( ) ( ) ( ) ( )( )( )

c cl c cl

cl

cl

N s N s D s sN s

s

+ ∆

=∆

• Regardless of the control topology used, the goal is to designa desired ( ), e.g., using Root Locus plots via thecl s∆

2c

Angle and M agnitude Conditions (or Coefficient M atching)to realize a desired time domain response and thentransform the associated analog controller ( ) to the z-domain,cG sg ( ) ,e.g., using the bilinear transformation, to create ( ); realizethe need to test the closed-loop responses of you

c

cG zr designs in both

the analog and digital domains due to the effect of the closedthe analog and digital domains due to the effect of the closed-loop zeros and approximations, e.g., associated with the bilineartransformation, on the various TFs, particularly ( )cG z

Common Controller Forms•

( )Some common controller forms are (1) lead:

( )( )

( ) ,leadc

lead

K s zG s

s p+

=+

where is to right of (leads) on the RL plot,(2) lag:

lead leadz p

( )( )

( ) ,lagc

K s zG s

s p

+=

+( )where

lags p+

is to left of (lags) on the RL plot,lag lagz p

( )( )and (3) lead-lag:

( ) lead lagK s z s zG s

+ +=

( )( ) ( )c

lead lag

G ss p s p

=+ +

Common Controller Forms• A proportional controller is a simple gain:

( ) ,

while a proportional-derivative (PD) controller is a specialc pG s K K= =

( )form of lead: ( ) ,c p d leadG s K K s K s z= + = + Note: plead is at infinity( )( )

a proportional-integral c p d lead

( )(PI) controller is a special form of lag:

K s zK +( ) ( ) ,

and a proportional integral derivative (PID) controller is a

lagic p

K s zKG s Ks s

+= + = Note: plag is at zero

( )

and a proportional-integral-derivative (PID) controller is aspecial form of lead-lag:

K ( )( ) ( ) leadi

c p d

K s z sKG s K K ss

+= + + =

( )lagzs

+

Controller Design Example #1•

( )1Given the plant: ( ) pG s =

( )G ve t e p a t: ( )

1design a lead cascade controller of the form:

pG ss s +

( )( )

( ) leadc

lead

K s zG s

s p+

=+( )

that places the desired closed-loop system poles at: 3.54 3.54cls j= − ±

2 ( ) 7.08 25.06(see prior design example for related time-domain specs )

cl

cl

j

s s s⇒∆ = + +

(see prior design example for related time-domain specs.)

Controller Design Example #1• It is clear from a basic RL plot of the plant that the desiredp p

closed poles cannot be achieved by gain alone and that theresulting loci must be biased to move further into the left-halfgplane (see prior example for RL plot of this plant)

Gi th b ti d th RL l t l it i•

( )

Given these observations and the RL plot rules, it isreasonable to propose a lead compensator of the form:

( )( )

1 ( )

7.08c

K sG s

s+

=+

• This results in the following Type 1 loop gain:K

( ) ( )

7.08KL s

s s=

+

Controller Design Example #1• The location of the controller pole should be obvious

3 4 3 4

pfrom the RL rules, but can be calculated from the MC as:

( )L s π∠ = −

( )3.54 3.54

3.54 3.54 3.54 3.54

( )cl

cl cl

s j

leads j s j

s

s s p

π

π=− +

=− + =− +⇒ −∠ −∠ + = −

•

7.08

The control

leadp⇒ =

ler gain can be calculated from the AC as:The control

3.54 3.54

ler gain can be calculated from the AC as:

( ) 1cls jK L s =− + =

3 54 3 54

1 25.06( )

s j

KL s

+

⇒ = =3.54 3.54cls j=− +

Design Verification Using Coefficient Matching • It has been shown that the desired closed-loop characteristic

2

polynominal (CP) is given by: ( ) 7.08 25.06cl s s s∆ = + +

• Based on the proposed lead compensator form, we can calculatethe re

cl

sulting loop gain as:

( )( ) ( ) ( )

g p g

1( ) 1

lead

l d l d

K s z KL ss p s s s s p

+= ⋅ =

+ + +

•( ) ( ) ( )1

If we select 1 as noted previously, this results in a closed-loop CP of the form (see prior controller topology analysis slide):

lead lead

lead

s p s s s s pz+ + +

= −

loop CP of the form (see prior controller topology analysis slide): ( ) 2 ( )

i i i h i ldcl lead leads s s p K s p s K∆ = + + = + +

• Comparing coefficients of the two CPs yields: 7.08 and 25.06leadp K= =

Calculation of the Steady‐State Error • The steady-state error to a unit step input is given by:

1 lim ( )1ss step t step p

e e tk→∞

= =+

where is the position error constant given by:

25.06

p

pk

( )0 0

25.06 lim ( ) lim7.08p s s

k L ss s→ →

= = = ∞+

1

•

1 01

Note: this result is due to the pole at zero in the loop gain!

ss stepp

ek

⇒ = =+

•

•

Note: this result is due to the pole at zero in the loop gain!

It can also be shown that is finite, yet non-zeross rampe

using the results of Chapter 3p

The Closed‐Loop TF Hcl(s) and Sample Period Selection• Based on the loop gain ( ) and the cascade control topologyL s

(see prior analysis), we have:( ) 25.06( ) Y s KH s = = =

( ) 2 ( )( ) 7.08 25.06

which agrees with the prior design specifications develop

cllead

H sR s s s z K s s

= = =+ + + +

ed

•

g p g p p

We can use the ( ) associated with ( ) to selecte ani t l t f th l t d di t ti t ll

cl cls H s∆

appropriate sample rate for the related discrete time controller2by noting from Chapter 3 that 50 100d s dTπω ω ω≤ = ≤

• Since the closed-loop 1d n

T

ω ω= − 2 3.54, an appropriateζ =

sample period for the desired closed-loop system dynamics isgiven by the interval 0.02 0.04T≤ ≤

Converting Gc(s) to Gc(z)• If we select 0.02, we can use the inverse of the bilinearT =

( )transform to calculate ( ) as shown:

25.06 1 23.64 23.17cG z

s z+ −( )( )

2 11 1100

1

5.06 23.64 23.17( ) ( )7.08 0.8678c c zs

T z zsz

s zG z G ss z

−=+ −=

+

= = =+ −

Note: This transformation is commonly performed using aCAD package, like the c2d() command in Matlab, as shownin the scripts posted on the class web site that solve each of theChapter 5 controller examples in the lecture notes using Matlab

Converting Gp(s) to GZAS(z)• If we select 0.02, we can use the techniques introduced inT =

Chapter 3 to convert ( ) to ( ) as shown:

( )1 1 1

p ZASG s G z

G sz z − − ( )

( )

2

( )1 1 1( )1

0 00019867 0 9934

cZAS

G sz zG zz s z s s

= = + Z Z

( )0.00019867 0.9934

( 1z

z+

=− ) ( 0.9802)

N Thi f i i l f d iz −

Note: This transformation is commonly performed using aCAD package, like the d2c() command in Matlab, as shownin the scripts posted on the class web site that solve each of theChapter 5 controller examples in the lecture notes using Matlab

Computing the Loop Gain L(z) and TF Hcl(z)• The loop gain ( ) is simply the product of ( ) and ( )c ZASL z G z G z

( )0.0046961 0.9934as shown ( ) ( ) ( ) ,

( 1) ( 0.8678)c ZAS

zL z G z G z

z z+

= =− −( ) ( )

while the closed-loop TF for the cascade control loop is( ) ( )( ) Y z L zH 0.0046961 ( 0.9934)z +( ) ( )( )( ) 1clH zR z

= =+ 2

( )( )

Note: Since there is no unstable pole/zero cancellation in ( )

( 1.863 0.8724)L z

L

z z− +=

Note: Since there is no unstable pole/zero cancellation in ( )the closed-loop system is internally stable, as discussed inCh t 4

L z

Chapter 4

Note: This above computations are commonly performed usinga CAD package like Matlab's feedback() command

Computing the Closed‐Loop Step Response• The closed-loop step response can now be calculated as

( ) ( ) ( ) ( ) using the inverse1cl clstep

zY z H z R z H zz

= = ⋅ − z-transform techniques discussed in Chapter 2 and compared to the desired closed-loop step response specifications

Note: This computation is commonly performed using aCAD package like the Matlab step() or residue() command;CAD package like the Matlab step() or residue() command;see the scripts posted on the class web site that solve each ofth Ch t 5 t ll l d i th l t tthe Chapter 5 controller examples covered in the lecture notes

Discrete Controller Implementation• The cascade controller is physically implemented in hardware

1

in the discrete time domain using the inverse-z TF as shown:( ) 23.64 23.17 23.64 23.17( ) U z z zG

−− −

( )

1

1

( )( )( ) 0.8678 1 0.8678

1 0 8678 ( ) 23

cG zE z z z

U

−

−

= = =− −

( )164 23 17 ( )E−( )11 0.8678 ( ) 23z U z⇒ − = ( )1.64 23.17 ( )

( ) 23.64 ( ) 23.17 ( 1) 0.8678 ( 1)

z E z

u k e k e k u k

−

= − − + −( ) ( ) ( ) ( )where ( ) is the control signal and ( ) is the error defined:

( ) ( ) ( )u k e k

e k r k y k= − ( ) ( ) ( )e k r k y k

Comments on the Remaining Examples• To save space, the controllers discussed in the remaining

examples will not cover all steps in the process of convertingan analog prototype controller into a discrete controller,but the calculation techniques used remain the same as illustrated in Example #1

•p

The remaining controllerswill be designed for thewill be designed for theanalog Mass-Spring-Dampersystem shown to the rightsystem shown to the rightrepresenting a simplifiedq arter car model:quarter car model:

Comments on the Remaining Examples• Analyzing the MSD system yields: ( ) ( ) ( ) ( )My t By t Ky t u t+ + =

2 21, 4, 3

( ) 1 1( )( ) 4 3p

M B K

Y sG sU s Ms Bs K s s= = =

= = =+ + + +

( )( ) ( ) ( )

, ,

1 1 2 1 2 1 3 1 3s s s s

= = −+ + + +

•

( )( ) ( ) ( )

This is a stable second order Type 0 plant consisting of twofirst order poles (modes) with the impulse response:fir

( )3

st order poles (modes) with the impulse response:1 ( )2

t tpg t e e− −= −

•

( )2

Clearly, this is a poor response for an auto suspension since itis overdamped with large time constants and has non zerois overdamped with large time constants and has non-zerosteady-state error to a step input, e.g., a bump in the road

Controller Design Example #2

• 1Given the plant: ( )G s =• ( )( )Given the plant: ( )

1 3design a lead cascade controller of the form:

pG ss s

=+ +

( )( )

g

( ) leadc

l d

K s zG s

s p+

=+( )

that places the desired closed-loop system poles at:leads p+

3 54 3 54s j= − ± 2

3.54 3.54

( ) 7.08 25.06( i d i l f l t d ti d i )

cl

cl

s j

s s s

= − ±

⇒ ∆ = + +(see prior design example for related time-domain specs.)




( )

Given these observations and the RL plot rules, it isreasonable to propose a lead compensator of the form:

( )( )

3 ( )

6.08c

K sG s

s+

=+

• This results in the following Type 0 loop gain:

( ) K( )( )

( )1 6.08KL s

s s=

+ +

Controller Design Example #2• The location of the controller pole should be obvious

3 4 3 4

pfrom the RL rules, but can be calculated from the MC as:

( )L s π∠ = −

( ) ( )3.54 3.54

3.54 3.54 3.54 3.54

( )

1cl

cl cl

s j

leads j s j

s

s s p

π

π=− +

=− + =− +⇒ −∠ + −∠ + = −

•

6.08

The contr

leadp⇒ =

oller gain can be calculated from the AC as:The contr

3.54 3.54

oller gain can be calculated from the AC as:

( ) 1cls jK L s =− + =

3 54 3 54

1 18.98( )

s j

KL s

+

⇒ = =3.54 3.54cls j=− +

Design Verification Using Coefficient Matching • It has been shown that the desired closed-loop characteristic

2

polynominal (CP) is given by: ( ) 7.08 25.06cl s s s∆ = + +•

( )Based on the proposed lead compensator form, we can calculatethe re

cl

sulting loop gain as:the re

( )( ) ( )( ) ( )( )

sulting loop gain as:

1( ) 1 3 1

leadK s z KL ss p s s s s p

+= ⋅ =

+ + + + +

•( ) ( )( ) ( )( )1 3 1

If we select 3 as noted above, this results in a closed-l CP f th f ( i t ll t l l i lid )

lead lead

lead

s p s s s s pz+ + + + +

= −loop CP of the form (see prior controller topology analysis slide): c∆ ( )( ) ( ) ( )2( ) 1 1l lead lead leads s s p K s p s p K= + + + = + + + +

( ) ( )1 7.08 6.08 and 25.06 18.98

Comparing the two CPs and solving simultaneous eqns. yields:

lead lead leadp p p K K+ = ⇒ = + = ⇒ =



e e tk→∞

= =+

where is the position error constant given by:

18 98

p

pk

( )( )0 0

18.98 lim ( ) lim1 6.08p s s

k L ss s→ →

= =+ +

3.12

1

=

1 0.241ss step

p

ek

⇒ = =+

• Unlike the previous example, which had a Type 1 loop gainand thus zero steady-state error, the loop gain of this exampleis Type 0, resulting in a finite (non-zero) error to a step inputand an infinite error to a ramp input



1 3design a lead-lag cascade controller of the form:

pG ss s

=+ +

( )( )( )( )

g g

( ) lead lagc

K s z s zG s

s p s p

+ +=

+ +( )( )that places the desired closed-loop system poles at:

lead lags p s p+ +

2

3.54 3.54

( ) 7.08 25.06cl

cl

s j

s s s

= − ±

⇒ ∆ = + +while producing zero steady-state error to a step input(see prior design example for related time-domain specs.)




( )( )

Given these observations and the RL plot rules, it isreasonable to propose a lead-lag compensator of the form:

( )( )( )

1 3 ( )

7.08c

K s sG s

s s+ +

=+

• This results in the same Type 1 loop gain of example #1,and therefore, the same gain 25.06 and error analysisK =and therefore, the same gain 25.06 and error analysisK



1 3design a PD cascade controller of the form:

pG ss s

=+ +

( )g

( )

that places the desired closed-loop system poles at:c p d leadG s K K s K s z= + = +

that places the desired closed loop system poles at: 5 5 7.07 and =0.707cl ns j ω ζ= − ± ⇒ =

2( ) 10 50s s s⇒∆ + + ( ) 10 50and compute the steady-state error to a step input

cl s s s⇒∆ = + +

•( )

Given the above, the resulting loop gain is:

( ) leadK s zL

+( )( )( )

( )1 3

leadL ss s

=+ +

Controller Design Example #4• The location of the controller zero can be calculated from the

( ) ( ) ( )5 5

5 5 5 5 5 5

MC as: ( )

1 3cls j

lead j j j

L s

s z s s

π

π=− +

∠ = −

⇒∠ + −∠ + −∠ + = −( ) ( ) ( )5 5 5 5 5 5

1 5tan 60.5 7.835

cl cl cl

lead

lead s j s j s j

z leadzθ

=− + =− + =− +

− ⇒ = = ⇒ =

•

5

The controller gain c

leadz leadleadz −

an be calculated from the AC as:

5 5( ) 1cls jK L s =− + =

( ) ( )( )

1 5 5 3 5 51 6.0( ) 7 83 5 5

j jK

L s j− + − +

⇒ = = =− +

•

( )5 5( ) 7.83 5 5

This implies that 46.9 and 6.0

cls j

p lead d

L s j

K K z K K

=− + − +

= ⋅ = = =

Graphical Interpretation of MC and AC

5.755.34

6.40

leadzθ 128111

180 111 8 128 7 60 5θ = + + =

5.34 6.406.0

5 75

180 111.8 128.7 60.5leadz

K

θ

⋅

= − + + =

= =5.75



e e tk→∞

= =+

( )where is the position error constant given by:

6.0 7.83pk

s +( )( )0 0

6.0 7.83 lim ( ) lim

1p s s

sk L s

s s→ →= =

+ +( )15.66

31

=

1 0.061ss step

p

ek

⇒ = =+

• Unlike the previous example, which had a Type 1 loop gainand thus zero steady-state error to a step, the loop gain of thisexample is Type 0, resulting in a finite (non-zero) error to astep and an infinite error to a ramp input



1 3design a PID cascade controller of the form:

pG ss s

=+ +

( )( )g

( ) lead lagic p d

K s z s zKG s K K ss s

+ += + + =

that places the desired closed-loop system poles at:5 5 7 07 and =0 707

s s

s j ω ζ= − ± ⇒ = 5 5 7.07 and 0.707

cl ns j ω ζ= ± ⇒ =

2 ( ) 10 50and ields ero stead state error to a step inp t

cl s s s⇒∆ = + +

and yields zero steady-state error to a step input


• The resulting loop gain is given by:•( )( )( )( )

The resulting loop gain is given by:

( ) 1 3lead lagK s z s z

L s+ +

=

•

( )( )( )

1 3

We can create a Type 1 loop gain that is similar to example #4

s s s+ +

yp p g pif we choose 1, which results in:lagz = −

( )K s z+ ( )L s ( )

( )3leadK s z

s s+

=+

• Using the same analysis techniques for example #4, we find that7.14 and 7.0, as shown on the next slide:leadz K= =

Controller Design Example #5• The location of the controller zero can be calculated from the

( ) ( )5 5

MC as: ( )

3cls j

l d

L s

s z s s

π

π=− +

∠ = −

⇒∠ + −∠ −∠ + = −( ) ( )5 55 5 5 5

1

3

5tan 66.8 7.14

clcl cllead s js j s j

z lead

s z s s

z

π

θ

=− +=− + =− +

−

⇒ ∠ + ∠ ∠ +

⇒ = = ⇒ =

•

5

The controller gain can

leadz leadleadz −

be calculated form the AC as:g

5 5( ) 1cls jK L s =− + =

( )( )5 5

5 5 3 5 51 7.0( ) 7.14 5 5j

j jK

L s j− + − +

⇒ = = =− +( )5 5

( )cls j j=− +

Controller Design Example #5• This implies that:

( )( ) 27.0 7.14 1( ) d p ic

K s K s Ks sG s

s s+ ++ +

= =

therefore, 56.98, 49.98 and 7.0p i d

s s

K K K= = =

• Due to the uncanceled pole at zero in the controller,the loop gain is Type 1 and thus the steady-state errorthe loop gain is Type 1 and thus the steady state errorto a step input is zero and finite (non-zero) to a ramp input

Empirical Tuning of PID Controllers• Assuming the plant can be modeled as a first order system

with delay (dead time) of the form:

( ) LsKG s e−=( )

( )1

plant parameters can be estimated using the Tangent Method,

pG s esτ

=+

as outlined below:

Note: Choose one of the two estimatesshown for the system time constant

Ziegler‐Nichols Tuning of P, PI, PID Controllers• Given the plant model parameters , , , e.g., using theK Lτp p g g

Tangent Method, you can tune PID controllers using thefollowing scheme designed to provide disturbance rejection:g g p j

Empirical Tuning Example • Assuming the plant step response (shown below) can be

modeled as a first-order system with delay, the TangentMethod yields the following estimated system parameters:

1, 1.55, 3K L τ= = =

Empirical Tuning of PID Controllers•Using Ziegler Nichols tuning with these system parameters−

results in the following PID gains 2.32, 0.75, 1.8

and resulting closed-loop step response w/step load disturbancep i dK K K= = =

g p p p p

Comment on the Remaining Material• The remaining material will not be covered on an exam, as it

shows how analog controllers are implemented in the analogdomain, i.e., using active or passive circuits; this would be discussed in more detail in a course like EE450/550

Active Compensator Realizations

Lead‐Lag Compensator w/Isolation

Passive Realization of Compensators

TF of Passive Lead‐Lag Compensator

ee451/551: digital control - web2.clarkson.edu · root locus method • it is desired to find the...

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